What is the predicted change in the boiling point of water when 2.10 g of

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Pani-rosa [81]

The question is incomplete, here is the complete question:

Calculate the solubility of hydrogen in water at an atmospheric pressure of 0.380 atm (a typical value at high altitude).

Atmospheric Gas Mole Fraction kH mol/(L*atm)

$N_2$ $7.81\times 10^{-1}$ $6.70\times 10^{-4}$

$O_2$ $2.10\times 10^{-1}$ $1.30\times 10^{-3}$

Ar $9.34\times 10^{-3}$ $1.40\times 10^{-3}$

$CO_2$ $3.33\times 10^{-4}$ $3.50\times 10^{-2}$

$CH_4$ $2.00\times 10^{-6}$ $1.40\times 10^{-3}$

$H_2$ $5.00\times 10^{-7}$ $7.80\times 10^{-4}$

Answer: The solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

Explanation:

To calculate the partial pressure of hydrogen gas, we use the equation given by Raoult's law, which is:

$p_{\text{hydrogen gas}}=p_T\times \chi_{\text{hydrogen gas}}$

where,

$p_A$ = partial pressure of hydrogen gas = ?

$p_T$ = total pressure = 0.380 atm

$\chi_A$ = mole fraction of hydrogen gas = $5.00\times 10^{-7}$

Putting values in above equation, we get:

$p_{\text{hydrogen gas}}=0.380\times 5.00\times 10^{-7}\\\\p_{\text{hydrogen gas}}=1.9\times 10^{-7}atm$

To calculate the molar solubility, we use the equation given by Henry's law, which is:

$C_{H_2}=K_H\times p_{H_2}$

where,

$K_H$ = Henry's constant = $7.80\times 10^{-4}mol/L.atm$

$p_{H_2}$ = partial pressure of hydrogen gas = $1.9\times 10^{-7}atm$

Putting values in above equation, we get:

$C_{H_2}=7.80\times 10^{-4}mol/L.atm\times 1.9\times 10^{-7}atm\\\\C_{CO_2}=1.48\times 10^{-10}M$

Hence, the solubility of hydrogen gas in water at given atmospheric pressure is $1.48\times 10^{-10}M$

4 0

3 years ago

How are viscosity and flow rate similar

Stels [109]

Answer: Flow rate is inversely proportional to viscosity.

Explanation: The relation is called Poiseuille's law, which describes the smooth flow of a fluid along a tube.

7 0

3 years ago

Read 2 more answers

Write a balanced net ionic equation for the following reaction: SrCl2(aq) + H2CO3(aq) → SrCO3(s) + HCl (aq)

max2010maxim [7]

We will balance the equation in the following order: metals, amethals, carbon, hydrogen and oxygen (the most common order).
The metal present in the equation is Sr, which is already balanced (there are 1 on each side of the equation).
The amethal present in the equation is Cl. There is 2 Cl in the left side and only one in the right side. So, we will multiply the quantity of the molecule that contains Cl by 2. Doing this, we'll obtain:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

Looking at the equation, we can see that it is now fully balanced. Hence, a balanced equation of the reaction is:

$SrCl_2_{(aq)}+H_2CO_3_{(aq)}\to SrCO_3_{(s)}+2HCl_{(aq)}$

5 0

2 years ago

The number of nitrogen atoms in one mole of nitrogen gas are...

9966 [12]

Explanation:

The number of nitrogen atoms in one mole of nitrogen gas are 6.02214179×1023 nitrogen atoms. 

Hope this helps...

3 0

2 years ago

A 7.36 g sample of copper is contaminated with a additional 0.51 g sample of zinc. suppose an atomic mass measurement was perfor

OleMash [197]

The average molecular weight of the mixture can be calculated using this formula:
MWav = x1MW1 + x2MW2

Where x is the mass fraction of the components of the mixture, in this case, copper (63.546 g/mol) and zinc (65.38 g/mol).

x1 = 7.36 / (7.36+0.51)=0.935
x2 = 0.51 / (7.36+0.51)=0.065

So,
MWav = 0.935(63.546) + 0.065(65.38) = 63.665 g/mol

3 0

3 years ago